TITLE: Prove that if $a$ divides both $2$ and $\sqrt{10}$ in $\mathbb{Z}[\sqrt{10}]$, then $a$ is a unit
QUESTION [5 upvotes]: Prove that if $a$ divides both $2$ and $\sqrt{10}$  in $\mathbb{Z}[\sqrt{10}]$, then $a$ is a unit. 
Further, show that you can't express $a$ as  $a = 2b + \sqrt{10} c$ where $b, c \in \mathbb{Z}[\sqrt{10}]$.
The only hint provided is that you should use the fact that  $A^{2} + 10B^{2} = 2, -2$ is insoluble.

REPLY [2 votes]: Somewhere in the book it should say something along the lines of: 

The norm function is defined thus: $N(A + B \sqrt{10}) = A^2 - 10B^2$.

It should also say something like:

This function is multiplicative.

These are worth reviewing; the book might even provide concrete examples that make the exercises much easier to do.
Check that $N(2) = 4$ and $N(\sqrt{10}) = -10$.
If $a|2$ and $a|\sqrt{10}$, then $2 = ak$ and $\sqrt{10} = aj$; we don't really care what $k$ and $j$ are but we do care that $k, j \in \mathbb{Z}[\sqrt{10}]$.
If $a$ is not a unit, then $|N(a)| \neq 1$ (the bars represent the absolute value function, in case you're wondering). Therefore $N(a) = 2$ and $N(k) = 2$ also ($N(a) = -2$ and $N(k) = -2$ works, too). But like the hint said, $|A^2 - 10B^2| = 2$ has no solutions, meaning that no number $x \in \mathbb{Z}[\sqrt{10}]$ satisfies $|N(x)| = 2$.
But since $a$ is not a unit, this means that $N(a) = 4$ and $N(k) = 1$ (so $k$ must be the unit). But now we have a much bigger problem when we try to make sense of $\sqrt{10} = aj$, since the only possible solution to $-10 = 4N(j)$ is $N(j) = -\frac{5}{2}$, but that takes us out of $\mathbb{Z}[\sqrt{10}]$ to the "wider world" of $\mathbb{Q}(\sqrt{10})$. Maybe $N(j) = 5$ and $N(a) = 2$, but, wait a minute, no, that doesn't work either.
Denying that $a$ is a unit has prevented us from figuring out $2 = ak$ and $\sqrt{10} = aj$. When you have eliminated the impossible... $|N(a)| = 1$ and there are infinitely many values of $a$ that also satisfy both $a|2$ and $a|\sqrt{10}$.